Adiabatic Process

Because the pump is insulated and the compression is sudden, no heat is exchanged, making this an adiabatic process with $Q=0$. The First Law written as $$\begin{gathered} \Delta \\ cup=Q-W \end{gathered}$$, where $W$ is the work done by the gas, gives $$\begin{gathered} \Delta \\ cup=-W_{\text{by gas}} \end{gathered}$$. Since 150 J of work is done on the gas, the work done by the gas is $-150\text{ J}$, so $$\begin{gathered} \Delta \\ cup=-(-150)=+150\text{ J} \end{gathered}$$. The decrease option mishandles the sign and would apply to adiabatic expansion where the gas does work. The unchanged option wrongly treats the process as isothermal, which cannot occur without heat flow during compression. The value 250 J invents extra energy that violates the energy balance. It is worth distinguishing the two sign conventions: some texts write the First Law as $$\begin{gathered} \Delta \\ cup=Q+W_{\text{on}} \end{gathered}$$, where work done on the gas is positive, which gives the same $+150\text{ J}$ directly. Either convention leads to the identical physical conclusion as long as it is applied consistently throughout the calculation. As a sanity check on signs, compressing a gas always raises its internal energy and temperature when no heat can leave, so a positive 150 J change is fully consistent with the physical picture.